Identify the molecule (X) with the maximum nu | vedclass

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(C) Let us calculate the total number of lone pairs for each molecule:$HNO_{3}$: $N$ has $0$, $O$ atoms have $2+2+3 = 7$ lone pairs. Total = $7$.$H_{2

Vedclass · Accepted Answer

$102$

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(C) Let us calculate the total number of lone pairs for each molecule:$HNO_{3}$: $N$ has $0$, $O$ atoms have $2+2+3 = 7$ lone pairs. Total = $7$.$H_{2}SO_{4}$: $S$ has $0$, $O$ atoms have $2+2+2+2 = 8$ lone pairs. Total = $8$.$NF_{3}$: $N$ has $1$, each $F$ has $3$. Total = $1 + 3(3) = 10$ lone pairs.$O_{3}$: Central $O$ has $1$, terminal $O$ atoms have $2$ and $3$. Total = $6$ lone pairs.Thus, $NF_{3}$ has the maximum number of lone pairs $(10)$.The hybridization of $N$ in $NF_{3}$ is $sp^{3}$. Due to the high electronegativity of $F$ atoms, the bond angle is compressed to approximately $102^{\circ}$.

Identify the molecule $(X)$ with the maximum number of lone pairs of electrons (obtained using Lewis dot structure) among $HNO_{3}$, $H_{2}SO_{4}$, $NF_{3}$, and $O_{3}$. Choose the correct bond angle made by the central atom of the molecule $(X)$. (in $^{\circ}$)

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